Problem: Find the equation whose graph is a parabola with vertex $(2,4)$, vertical axis of symmetry, and contains the point $(1,1)$. Express your answer in the form "$ax^2+bx+c$".
Since the axis of symmetry is vertical and the vertex is $(2,4)$, the parabola may also be written as  \[y=a(x-2)^2+4\] for some value of $a$.  Plugging the point $(1,1)$ into this expression gives  \[1=a(1-2)^2+4=a+4.\] This tells us $a=-3$.

Our equation is  \[y=-3(x-2)^2+4.\] Putting it $y=ax^2+bx+c$ form requires expanding the square, so we get  \[y=-3(x^2-4x+4)+4=\boxed{-3x^2+12x-8}.\]